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Chapter 21: Problem 39

Give formulas for the following. a. potassium tetrachlorocobaltate(II) b. aquatricarbonylplatinum(II) bromide c. sodium dicyanobis(oxalato)ferrate(III) d. triamminechloroethylenediaminechromium(III) iodide

Short Answer

Expert verified
The short answer for the formulas of the coordination compounds is: a. Potassium tetrachlorocobaltate(II): \(K2[CoCl4]\) b. Aquatricarbonylplatinum(II) bromide: \([Pt(H2O)(CO)3]Br2\) c. Sodium dicyanobis(oxalato)ferrate(III): \(Na3[Fe(CN)2(C2O4)2]\) d. Triamminechloroethylenediaminechromium(III) iodide: \([Cr(NH3)3Cl(en)]I3\)

Step by step solution

01

a. Potassium tetrachlorocobaltate(II)

In this coordination compound, the central metal atom is cobalt (Co) with an oxidation state of +2 (indicated by II in the name). The ligands are four chloride ions (Cl-), and the counterion is potassium (K+). So the formula for potassium tetrachlorocobaltate(II) is: K2[CoCl4]
02

b. Aquatricarbonylplatinum(II) bromide

In this compound, the central metal atom is platinum (Pt) with an oxidation state of +2 (indicated by II in the name). The ligands are one water molecule (H2O) and three carbonyl ligands (CO). The counterion is bromide (Br-). So the formula for aquatricarbonylplatinum(II) bromide is: [Pt(H2O)(CO)3]Br2
03

c. Sodium dicyanobis(oxalato)ferrate(III)

In this compound, the central metal atom is iron (Fe) with an oxidation state of +3 (indicated by III in the name). The ligands are two cyanide ions (CN-) and two oxalato ligands (C2O4虏-). The counterion is sodium (Na+). So the formula for sodium dicyanobis(oxalato)ferrate(III) is: Na3[Fe(CN)2(C2O4)2]
04

d. Triamminechloroethylenediaminechromium(III) iodide

In this compound, the central metal atom is chromium (Cr) with an oxidation state of +3 (indicated by III in the name). The ligands are three ammonia molecules (NH3), one chloride ion (Cl-), and one ethylenediamine ligand (en, C2H4(NH2)2). The counterion is iodide (I-). So the formula for triamminechloroethylenediaminechromium(III) iodide is: [Cr(NH3)3Cl(en)]I3

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Most popular questions from this chapter

When concentrated hydrochloric acid is added to a red solution containing the \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) complex ion, the solution turns blue as the tetrahedral \(\mathrm{CoCl}_{4}^{2-}\) complex ion forms. Explain this color change.

Ethylenediaminetetraacetate (EDTA \(^{4-} )\) is used as a complexing agent in chemical analysis with the structure shown in Fig. \(21.7 .\) Solutions of EDTA \(^{4-}\) are used to treat heavy metal poisoning by removing the heavy metal in the form of a soluble complex ion. The complex ion virtually prevents the heavy metal ions from reacting with biochemical systems. The reaction of EDTA \(^{4-}\) with \(\mathrm{Pb}^{2+}\) is $$\mathrm{Pb}^{2+}(a q)+\mathrm{EDTA}^{4-(a q)} \rightleftharpoons \mathrm{PbEDTA}^{2-}(a q) \\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad K=1.1 \times 10^{18}$$ Consider a solution with 0.010 mol of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) added to 1.0 \(\mathrm{L}\) of an aqueous solution buffered at \(\mathrm{pH}=13.00\) and containing 0.050\(M \mathrm{Na}_{4} \mathrm{EDTA} .\) Does \(\mathrm{Pb}(\mathrm{OH})_{2}\) precipitate from this solution? \(\left[K_{\mathrm{sp}} \text { for } \mathrm{Pb}(\mathrm{OH})_{2}=1.2 \times 10^{-15}\right]\)

Oxalic acid is often used to remove rust stains. What properties of oxalic acid allow it to do this?

Compounds of copper(II) are generally colored, but compounds of copper(I) are not. Explain. Would you expect \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) to be colored? Explain.

Draw all geometrical and linkage isomers of \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{NO}_{2}\right)_{2}\)
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